What is the sum from n=1 to infinity of 7/(n^2+n) ?

The sum from n=1 to infinity of 7/(n^2+n) is 7

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sum from n=1 to infinity of 7/(n^2+n)

n = 1 \sum \infty \frac{7}{n ^{2} + n}

7

Solution steps

n = 1 \sum \infty \frac{7}{n ^{2} + n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 7 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + n}

Apply Telescoping Series Test:

1

= 7 \cdot 1

Simplify

= 7

i = 1 \sum \infty 5 \cdot i

n = 1 \sum \infty n^{- 1 - \frac{1}{n}}

n = 1 \sum \infty - \frac{1}{1 5 ^{n}}

n = 1 \sum \infty \frac{n + 4}{4 ^{n}}

n = 0 \sum \infty e^{- 21 n}

What is the sum from n=1 to infinity of 7/(n^2+n) ?
The sum from n=1 to infinity of 7/(n^2+n) is 7